两个n位二进制整数相加问题

（原问题见算法导论第三版2.1-4）
问题：考虑把两个n位二进制数加起来的问题。这两个整数分别存储在两个n元数组A和B中。这两个整数的和应按二进制形式存储在一个（n + 1）元数组C中。
解答：
代码如下。

#include <stdio.h>#include <stdlib.h>#include <string.h>void binary_add(int *, int *, int *, int);int main(int argc, char const *argv[]){    /* code */    if (argc < 3) {        fprintf(stderr, "please input enough binary number\n");        return -1;    }    const char* num1 = argv[1];    const char* num2 = argv[2];    int n1 = strlen(num1);    int n2 = strlen(num2);    int n = (n1 > n2) ? n1 : n2;    int * left = calloc(n1, sizeof(int));    int * right = calloc(n2, sizeof(int));    int * result = calloc(n + 1, sizeof(int));    for (int i = 0; i < n1; ++i) {        if (num1[i] == '1') {            left[n1 - i - 1] = 1;        } else if (num1[i] == '0') {            left[n1 - i - 1] = 0;        } else {            fprintf(stderr, "error input left number, please input binary number\n");            return -1;        }    }    for (int i = 0; i < n2; ++i) {        if (num2[i] == '1') {            right[n2 - i - 1] = 1;        } else if (num2[i] == '0') {            right[n2 - i - 1] = 0;        } else {            fprintf(stderr, "error input right number, please input binary number\n");            return -1;        }    }    for (int i = 0; i < n + 1; ++i)    {        if (i < n + 1 - n1) {            printf(" ");        } else {            printf("%d", left[n1 - (i - n - 1 + n1) - 1]);        }    }    printf("\n");    for (int i = 0; i < n + 1; ++i)    {        if (i < n + 1 - n2) {            printf(" ");        } else {            printf("%d", right[n2 - (i - n - 1 + n2) - 1]);        }    }    printf("\n");    binary_add(left, right, result, n);    for (int i = 0; i < n + 1; ++i)    {        printf("%d", result[n - i]);    }    printf("\n");    return 0;}void binary_add(int *left, int *right, int *result, int n) {    int c = 0;    int temp;    for (int i = 0; i < n; i++) {        temp = left[i] + right[i] + c;        result[i] = temp % 2;        c = temp / 2;    }    if (c != 0) {        result[n] = c;    }}